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# Writing exponential functions from graphs

Given a graph of a line, we can write a linear function in the form y=mx+b by identifying the slope (m) and y-intercept (b) in the graph. GIven a graph of an exponential curve, we can write an exponential function in the form y=ab^x by identifying the common ratio (b) and y-intercept (a) in the graph. Created by Sal Khan.

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• general form of y= r . a^x is never really introduced--it just sort of appears in the context of problem videos. i found this confusing as i had not seen this before • Actually, the general form of an exponential function isn't f(x)= a*r^x, which is why I believe that it is not formally introduced in these videos. Really, the question just wants you to format your answer that way. And to be quite honest, f(x)=a*r^x is the most logical way to format your answer. The general equation for an exponential function is f(x)=ab^(x-h) +k not f(x)=a*r^x. I hope that answers your question!
• At , couldn't he just find the y-intercept using the graph? • Yes he can, in this case it would have been very straightforward. for f(x), (0,5) and for g(x), (0,3). Then you can use point (1,1) (or the other point, but (1,1) is simpler) to find m and r respectively.

I think the intent was to show another way of doing it, as what you suggested was shown in a previous video
• Am I the only one who noticed the video cut before he could finish? • At couldn't Sal have just looked at the graph of f(x) to see that the y-intercept (b) was 5? This won't be true in every case, but when the graph passes directly through a number that is marked on the graph you can read the graph at that number. Right? • I'm totally lost from about on. How does ar to the -1 become a/r? • How did r^2=1/9 become r=1/3 ? • why does sal always calculate the slope first? is there a reason for it (like, maybe it's necessary to do it that way in later math), or is it just a habit? because i usually look at the y intercept, then plug in a random set of points on the line to find the slope that way. i always mess up (proper) slope calculations for some reason, so doing it this way has a higher success rate with me, but if there's a reason to favor this method over my own then i'd like to know so i can get used to it sooner. • I have a question. I looked at this video and learned how to find the equation of exponential functions, but how would you find an exponential function if it had dilations, vertical translations, and horizontal translations? • All functions basically react the same to dilations and translations. The most basic exponential function is y=2^x. Looking at some of the key features, you have no x intercept, a y intercept at 1 (when x=0, 2^0=1) (0,1), then additional points of (1,2)(2,4)(3,8)(4,16)(-1,1/2)(-2,-1/4) etc. The horizontal asymptote is at y=0. So you could start adding changes to this such as y=a 2^(x-b)+c. If a is negative, it reflects across x. If a is >1, it multiplies all x values by a (for example if a=2, you would get (-2,1/2)(-1,1)(0,2)(1,4)(2,8). If a<1, then it cuts the points by the scale factor. If b is negative, it shifts to right, and if b is positive it shifts left, if c is negative it moves down and if c is positive, it moves up (the asymptote is moved up or down c units).  • Well, for a graph like that, you could just look at when `x = 0`, it is 3. So `a = 3`. Then you just pick a point for example, that point (1, 1). You will now have `1 = 3 * r^1` which is the same as `1 = 3r`. You can divide both sides by 3 which leaves you with `1/3 = r` saving quite a lot of space.